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In Fitzpatrick's Newtonian Dynamics book on the Coriolis force, he states

\begin{align} v_{x'}&\simeq V_0\cos\theta-2\Omega t V_0\sin\lambda~\sin\theta \tag{433}\\ v_{y'}&\simeq-V_0\sin\theta-2\Omega t V_0\sin\lambda~\cos\theta \tag{434} \end{align} To lowest order in $\Omega$, the above equations are equivalent to \begin{align} v_{x'}&\simeq V_0\cos\left(\theta+2\Omega t\sin\lambda \right)\tag{435}\\ v_{y'}&\simeq -V_0\sin\left(\theta+2\Omega t\sin\lambda \right)\tag{436} \end{align}

I simply don't get how the above equations are equivalent. In fact, if I take a magnitude (of the total vector) they are different. I also don't understand what "To lowest order in $\Omega$" means here.

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Note that Fitzpatrick states towards the beginning,

The following solution method exploits the fact that the Coriolis force is much smaller in magnitude that the force of gravity: hence, $\Omega$ can be treated as a small parameter

Generally, when statements like that are made, powers (greater than 1) of the term in question are considered to be zero: $$ \Omega\ll1; \quad \Omega^n\,\approx0\,\,\forall\,\,n>1 $$ Thus, $$ a\Omega+b\Omega^2\approx a\Omega $$ for small $\Omega$.

In the small angle approximation, we have \begin{align} \sin\theta&\approx\theta\\ \cos\theta&\approx1-\frac{\theta^2}2\\ \tan\theta&\approx\theta \end{align} We can apply this to your first $v_{x'}$ equation: \begin{align} v_{x'}&\approx V_0\left(\cos\theta-2\Omega\sin\lambda \sin\theta t\right)\\ &\approx V_0\left(1-\frac{\theta^2}2-2\Omega\sin\lambda\theta t\right) \end{align} You can then complete the square to finish the solution, ignore the $\Omega^2$ term and end up with Equation (435) that the author gives.

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  • $\begingroup$ I've completed the square and get $Vo((1-\Omega\theta sin\lambda t )^2 -\frac{\theta^2}{2})$ I cant come up with the final equation, hehe. $\endgroup$ – DLV Nov 27 '14 at 4:10
  • $\begingroup$ You should only be considering the terms with respect to $\theta$, so complete the square with $\theta^2/2+2\Omega\sin\lambda\theta t$. $\endgroup$ – Kyle Kanos Nov 27 '14 at 4:16
  • $\begingroup$ Wow. I'm just impressed. Is this a famous derivation or something? When does this stop being valid? How much time has to pass? Thanks. $\endgroup$ – DLV Nov 27 '14 at 4:30
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    $\begingroup$ Famous? Probably not, I cheated and expanded (435) and recognized the difference between the two relations as a completing the square :). It stops being a valid relation when the second term is no longer small (either $\Omega\approx1$ or $t$ being some nominal value). $\endgroup$ – Kyle Kanos Nov 27 '14 at 4:33
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Ahh, Richard Fitzpatrick. Great guy.
Ok, If you start with the second set of expressions, use the appropriate double-angle-formula and then assume the "angle" $2\Omega \sin\lambda t$ is small (note that the $t$ is not within the sin function!), you get the first expressions, e.g.

$$\cos(\theta+\phi) = \cos\theta\cos\phi - \sin\theta\sin\phi,$$ and then for small $\phi$, use the first-order terms in the Taylor expansion for the trig functions, i.e. $\cos\phi \simeq 1$, and $\sin\phi\simeq \phi$. In your case, $\phi = 2\Omega\sin\lambda t$, where again the $t$ is intended as a (linear) multiplicative factor and not inside the $\sin$ function.

Similarly, $$\sin(\theta+\phi) = \sin\theta\cos\phi + \cos\theta\sin\phi,$$ and you can take if from there...;-)

Woops, looks like Kyle beat me to the punch while I was typing this in!

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  • $\begingroup$ Out of curiosity, could Fitzpatrick's books become non-free any time? They are too good. I guess I'll download them for fear of this. Hehe. $\endgroup$ – DLV Nov 27 '14 at 6:50

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